Limit of (1-cos(x))/x as x approaches 0 | Derivative rules | AP Calculus AB | Khan Academy

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Showing that the limit of (1-cos(x))/x as x approaches 0 is equal to 0. This will be useful for proving the derivative of sin(x).

AP Calculus AB on Khan Academy: Bill Scott uses Khan Academy to teach AP Calculus at Phillips Academy in Andover, Massachusetts, and heÕs part of the teaching team that helped develop Khan AcademyÕs AP lessons. Phillips Academy was one of the first schools to teach AP nearly 60 years ago.

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sin x / ( 1 + cos x ) = tan ( x / 2 )

so you can also evaluate it as : lim ( x -- > 0 ) sin x / x * lim ( x -- > 0 ) [ sin x / ( 1 + cos x ) ]

= lim ( x -- > 0 ) tan ( x / 2 )

= tan 0

= 0

michaelempeigne

Nice crosshair, mind sharing the settings? Trying to improve my aim, thanks

mnasarrr

That exactly is the definition of the derivative at point 0 of cos(x)

lim cos(0) - cos(x) / 0 - x = d/dx cos(0) = sin(0) = 0
x->0

welovfree

is it possible to deduce that the limit as pi/2 - theta goes to zero of cos x/x is equal to 1 ?

jaccuzeaverroes

Hmmm we could have used limited development and find that when x->0 this expression is equal to x/2->0
But I didn't think about doing it like this ! Thank you for the idea !

leobourquin

Is the product of limits always equal to the limit of products?

PV

mr. khan, i literally owe you my degree

wrongplanet

lim(x>infinity) [sin x÷(x + cos x)] =?

mathkinginfinity

You could use L'Hopital's rule instead, it is way easier

maorus

So the limit when x tends to 0 of 1+ cos x : x is also 0?

felipegonzalezlelong

pls can somebody tell me why 1- cos^2 x = sin^2 x

mathworld

Limit of (1-cos(x))/x as x approaches 0 | Derivative rules | AP Calculus AB | Khan Academy

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